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Old 05-31-2010, 11:51 AM   #1 (permalink)
Strange Famous
follower of the child's crusade?
 
Monty Hall math problem

A friend mentioned this to me, and we have had a discussion on it. To give the background, here is a link to an extremely patronising wikipedia article about it.

Monty Hall problem - Wikipedia, the free encyclopedia

The basic setup is:

Quote:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? (Whitaker 1990)

Although not explicitly stated in this version, the solution is usually based on the additional assumptions that the car is initially equally likely to be behind each door and that the host must open a door showing a goat, must randomly choose which door to open if both hide goats, and must make the offer to switch. As the player cannot be certain which of the two remaining unopened doors is the winning door, most people assume that each of these doors has an equal probability and conclude that switching does not matter. In fact, under the standard assumptions the player should switch—doing so doubles the probability of winning the car, from 1/3 to 2/3.
In my opinion this is completely wrong.

If you are in the same scenario but you dont make any guess and they open box 3 and there is no prize there, the possibility that it is 1 or 2 is equal by any acount

Given that what you guess has no physical effect on whether the prizes is in 1 or 2, how can the fact you say "1" alter the probability of where the prize actually is? The phyical reality is not affected in any way by what you guess.

If this was acted out in actual life 99 times, then the car would be in box 1 33 times, box 2 33 times, and box 3 33 times (on average). Whether you change your choice or not, the box has the same possibility to be in either of the remaining boxes. Whatever the person chooses and whatever the show's host does, it makes no difference.
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